Blaga P. Lectures on the differential geometry of - tiera.ru

146 Chapter 4. General theory of surfaces
Example. For the helicoid
⎧
x = u cos v
⎪⎨
y = u sin v
⎪⎩ z = bv
we can define the orientation by putting
n(u, v) =
, (u, v) ∈ R 2 , b > 0,
�
b sin v b cos v
√ , − √
b2 + u2 b2 + u2 ,
On gets, after a straightforward computation:
�
1
G =
0
0
b2 + u2 �
,
⎛
H = ⎜⎝
0 b
0
b
√ b 2 +u 2
√ b 2 +u 2
⎞
⎟⎠ , A =
�
u
√ .
b2 + u2 ⎛
⎜⎝
0
b
(b 2 +u 2 ) 3/2
b
(b2 +u2 ) 3/2
4.12 The second fundamental form of an oriented surface
Definition 4.12.1. The second fundamental form of an oriented surface S is a map associating
to each a ∈ S the application ϕ2(a) : TaS × TaS → R given by
ϕ2(ξ, η) = −ϕ1(A(ξ), η), ∀ξ, η ∈ TaS. (4.12.1)
Remark. The minus sign in the previous definition is a consequence of our particular
choice of sign in the definition of the shape operator. We found natural to choose the
shape operator to be the differential of the spherical map rather then the opposite of the
differential, but then in the definition of the second fundamental form we had to introduce
an extra minus sign, in order to be consistent with the generally accepted definition of
the second fundamental form.
Proposition 4.12.1. For each a ∈ S , ϕ2(a) is a symmetrical bilinear form.
Proof. We take two arbitrary tangent vectors ξ, η ∈ TaS and two arbitrary real numbers
α, β ∈ R. Then we have, first of all:
ϕ2(η, ξ) = −ϕ1(A(η), ξ)
A
=
self-adjoint −ϕ1(η,
ϕ1
A(ξ)) =
symmetrical −ϕ1(A(ξ), η) = ϕ2(ξ, η),
which means that ϕ2 is symmetrical. Due to the symmetry, it is enough to prove the
linearity in the first variable only. We have
ϕ2(αξ1 + βξ2, η) = −ϕ1(A(αξ1 + βξ2), η)
A
=
linear −ϕ1(αA(ξ1) + βA(ξ2), η)
0
⎞
ϕ1
=
bilinear
ϕ1
=
bilinear −αϕ1(A(ξ1), η) − βϕ1(A(ξ2), η) = αϕ2(ξ1, η) + βϕ2(ξ2, η),
⎟⎠ .